what battery current is good for motors?

[JamesG]

Hi again!

This doesn't really relate to the motor calculations, but I figured it would be better to put it here anyway.

I've been looking into different types of batteries and I've noticed that most AA batteries have a maximum discharge rate of 2.2 Amps or so. I've been wondering if this is an important figure to worry about as the motors I'll be wanting to drive could demand a stall current of 5 A each.

Does that mean that I should try and get batteries which support around 10A maximum current draw, or can the problem be solved by using a fat capacitor as a "secondary" current source for the motors. Would the fact that I want to control the motors with a PWM signal change anything with wanting to use a Cap? I'm guessing not, but you never know

I'm assuming that the motors will in most cases NOT be drawing anywhere near 2.2A average current, if that were the case, what would happen? Would the motors just not have enough torque to do anything, would the batteries overheat or something like that?

Thanks

James

[Admin]

There are two values you need to be concerned about -> average current draw, and maximum current draw.

Look at the datasheets of all the parts you plan on using, and sum up the current draw. Then select a battery capable of meeting both of those values.

And its always best to round up, just in case you decide to add more stuff later in the future, or want to push the limits.


ps - I split this topic, t'was entirely unrelated to the previous thread

[JamesG]

There are two values you need to be concerned about -> average current draw, and maximum current draw.

Look at the datasheets of all the parts you plan on using, and sum up the current draw. Then select a battery capable of meeting both of those values.

And its always best to round up, just in case you decide to add more stuff later in the future, or want to push the limits.


ps - I split this topic, t'was entirely unrelated to the previous thread

Thanks for the reply admin, but I don't really feel much more knowledgeable on the topic.

I was pretty sure that that would be the method that one *should* use to calculate the required average and max current draw, but maybe I didn't make my actual question clear enough.

Assuming that I calculate the average required current draw, and it's below the average current provided by the battery, but the peak required current is much higher than the peak current that can be provided from the battery.

I know that in my robot, the peak current draw is mostly made up of the 10A of the two motors, but I also know/assume that in the operating environment of my robot it is unlikely that it will ever need to actually draw those 10A. How does this affect my circuit/plans?

Can I say that having a large ceramic cap in parallel with the battery will be enough to provide me with the instantaneous required current for the motors at any given point in time, assuming that the demand for high current will only last a very short period of time? If the demand for this high current were to last longer than the time that the cap could provide the current for, what would happen? I'm planning on having my MCU and motors running on the same battery pack, but through different, optoisolated parallel, circuits. How would the motors drawing a large amount of current affect the current provided to the MCU? If the circuits were parallel then it shouldn't have an affect on the voltage to the circuit, but could limit the current spectacularly, which I assume would also result in a reset of sorts.

Thanks for any further insight into this.

James

[waltr]

Good question on the battery specs but the answer is not easy. So I'll give it a try.

All batteries have what is known as internal resistance. This can be modeled as a resistor in series with the battery. An 'Ideal' battery has zero internal resistance has therefore has the ability to supply infinite current into a short circuit. It is this resistance that limits the maximum instantaneous current that can be drawn from a battery, the 2.2A for the AA batteries. The battery chemistry and construction determines what the internal resistance will be and thus what the maximum instantaneous current can be.

I take it your motors have a stall current of 5A (10A for two motors). When you start the motor from zero speed by applying the full voltage the instantaneous current draw can approach 5A but will decrease as the motor comes up the speed. If you are using PWM to control the motor speed then there will be higher current drawn each time the PWM switches the motor on.
Lets say you are using the AA batteries you asked about and both the motor and processor are connected to the same battery. When the motor turns on (PWM pulse) the higher current draw will cause the voltage at the battery to drop. How much is determined by the batteries internal resistance and Ohm's law. Vdrop = I * Rbat

If the current or internal resistance is high enough then the voltage drop can reset the processor, called brown-out.
A few large capacitors at the motor drive and at the processor can help supply the instantaneous current the motor requires and keep the voltage to both the motor and processor from dropping too much. But the capacitors can only do this for a short period of time as they will discharge while current is being drawn from them.

If the high instantaneous current is for a very short time then caps will work well to maintain the voltage in the circuits. However, if the high current draw is for an extended time, like when a motor is stalled, then the caps will discharge and the voltage to the circuits will drop.

There are several ways to work around this problem in addition to strategically place caps.
1- chose a battery that can always supply the maximum current the system may draw.
2- current limit the motors to under the maximum current the battery can supply
3- run the processor on a separate battery
4- design the motor, gears, wheels so that the motor never stalls and thus never draws the maximum current

Is this explanation what you are looking for?

[Soeren]

Hi,

Good question on the battery specs but the answer is not easy. So I'll give it a try.

All batteries have what is known as internal resistance. This can be modeled as a resistor in series with the battery. An 'Ideal' battery has zero internal resistance has therefore has the ability to supply infinite current into a short circuit. It is this resistance that limits the maximum instantaneous current that can be drawn from a battery, the 2.2A for the AA batteries.

You're riding on with the false assumption made by JamesG about the max. current.

[...]I've noticed that most AA batteries have a maximum discharge rate of 2.2 Amps or so

And then wander off into explaining it with Ri, which you obviously need to read up on (if you aren't 100% knowledged in a subject, better say so than trying to sound like you know - and by that bring false "knowledge" to others that might later spread that virally).

Most AA have a capacity of 2.2Ah (Ampere-hours) or so would have been a correct statement (with some levy to the actual capacity going to 2.9Ah for AAs these days).

It's still the consumption of the load that determines the current as long as it's under a certain amount.

To quote a Panasonic paper: "Since the internal resistance of NiMH is low, continuous high-rate discharges of up to 3C is possible, similar to NiCd"
The operative words here are "continuous high-rate discharges".
A NiMH will easily deliver short term currents several times that, so please forget about Ri and all that for a moment and let's give JamesG an answer that he can use - it's actually quite easy.

JamesG <- Your 2.2Ah cells are capable of more than the stall current, so do include a fuse or other protection (most motors will quickly die if kept stalled with the power on).
Aside from that, if your vehicles total (average) consumption is say 4.4A, a 2.2Ah battery will drive them for about 20..25 minutes.

The reason: 2.2Ah / 4.4A = 0.5h, but since the spec'd capacity of 2.2Ah is measured at either C5 or C20 (the 5 hour or 20 hour rate), with C5 as the most likely and the fact that capacity decreases with a rising current, it will be somewhat less than half an hour. The exact amount of time you get will depend on the quality of the cells, how old they are and how well you kept them.

In case of really high inrush currents, a huge capacitor can be used as a buffer or a slow start circuit can be added (which is best depends on what's hooked up to it).
In your case, just use as is.
Max. draw will happen, not when stalled, but in a short pulse when going directly from one direction to the opposite - if that seems a blow to the gut of your batteries, use a large cap.

No cap of a reasonable size (i.e. smaller than the vehicle itself) will be able to hold the power to the motor from dropping some shortly). That may well interfere with the controller, but that is easily taken care of by a diode and a capacitor, as the current draw is much lower on the controller side of things.
The formula for sizing the capacitor  is:
C = As/V
Where:
C = Capacitance in Farad
A = current consumed
s = seconds you need to hold the voltage
V = the magnitude of voltage that you will allow it to drop during s

Example.
If you need to keep the voltage drop to 1.5V max in 0.03s with a current draw of 0.5A, it will be:
C = 0.5 x 0.03 / 1.5 = 0.01F = 10,000 µF

With PWM, and a good choice of frequency, the current will be lower, as the induction of the motor will tend to integrate the voltage. A small cap here can be good or bad, depending on whether you hit close to resonance.

[waltr]

Ok Søren,
 I do need to read some more about this. I didn't realize that the original question was flawed, 2.2A max current draw verse 2.2AHr capacity and not stating what the battery chemistry was.  I was sure you would jump in and help out with the correct answer. Thanks.

[JamesG]

Soeren, thanks for providing so much useful info.

The confusion as to the peak output current of the batteries came from the website I was looking at, which had in places specified from 2-10A depending on the package of the cell (from AA-D). Also, you refer to "high-rate discharges of 2C", is that two times the battery capacity, or is it two Coloumbs?

I think I'm going to just use a fuse for the moment, I'm thinking that getting two fuses rated for about 2-3 A but then with a long enough blow-time so that they'll be ok for small peak currents, but then with a peak current for 1-2s of 3-5A they'll blow.

If I later feel the need to change that configuration I think I can get a current sensor or two and use the Arduino that I'm thinking of getting to throttle the output signal accordingly.

I'm inquisitive as to the exact workings of your last sentence:

With PWM, and a good choice of frequency, the current will be lower, as the induction of the motor will tend to integrate the voltage.

I'm a second year EE student, so I can begin to understand the implications of this, but not to the extent that I totally understand how it functions, could you possibly explain it a little?

Thanks,

James

[Admin]

Just to add on to what Soeren said about caps . . . its a very time-dependent function.

You need to know exactly how long peak current draw occurs to reliably calculate any capacitor value. And if you choose a value too large, when turning your robot on the voltage could increase so slowly that your mcu might not even start up. Basically, don't count on caps powering anything but your mcu.

Just to give you an idea . . . the big blue cap on the original Axon only has enough power to run an LED for ~3 seconds.

[Soeren]

Hi,

[...] "high-rate discharges of 2C", is that two times the battery capacity, or is it two Coloumbs?

Actually, it was 3C
And yes, the "C" is the capacity, adding a number in front (left) of the "C" it is a multiplier and a number to the right is a divisor.
So, 3C is 3 times the capacity [in Ah, or mAh] while C3 would be a third of the capacity.
For a 2.2Ah cell, 3C is then 6.6A and C3 is 733mA.

I'm inquisitive as to the exact workings of your last sentence:

With PWM, and a good choice of frequency, the current will be lower, as the induction of the motor will tend to integrate the voltage.

I'm a second year EE student, so I can begin to understand the implications of this, but not to the extent that I totally understand how it functions, could you possibly explain it a little?

PWM pulses are (ideally) pulses with a very fast rise time, to make it be either on or off, while keeping the transition through the linear area of the power stage short, as it's here that most power is dissipated.
For a brief moment, think of the motor as an induction. The current won't follow the fast rise time, as the induction will resist the build up, hence coils are very good for noise filters, but here we want the "noise".

LC circuits are even better noise filters, but as you know, an LC circuit has got a resonance frequency where things behave differently and can work as an "amplifier".

Unfortunately, there's no way of knowing beforehand how a certain motor will behave (unless you get real expensive and proper spec'd motors), but luckily, it's not that hard to find empirically.

The PWM frequency will be a compromise between how much whine you'll accept (loosely wound motors can be a real pain, like one of my cordless drills, that really ticks me off - and my hearing from 4kHz and above is 65dB down) and how efficient you want it.

If you go too high in frequency, the induction will really ruin the efficiency, as it smashes the rise time of the PWM. Going to low, audible screaming and, if really low, jerky motion are lurking.'
Usually, you'll have to accept a bit of a whine, but with a good motor, it should be reasonably soft.

Best way is to experiment and using a 'scope on different constellations will teach you faster than reading about it

Did any of these ramblings answer your question?